Circle Questions for RRB NTPC PDF
Download RRB NTPC Top-15 Circle Questions PDF. Questions based on asked questions in previous exam papers very important for the Railway NTPC exam.
Download Circle Questions for RRB NTPC PDF
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Question 1: A sector of a circle subtending angle 36° has an area 3.85 $cm^{2}$. The length of the arc is
a) 2.2 cm
b) 2.5 cm
c) 2.7 cm
d) None of these
Question 2: A wire when bent in the form of a square encloses an area of 484 sq. cm. If the same wire is bent in the form of a circle, what is the area enclosed by it?
a) 264 sq. cm
b) 616 sq. cm
c) 488 sq. cm
d) 492 sq. cm
Question 3: If the circumference of a circle is 22 cm, find the area of the semicircle.
a) 38.5 sq.cm
b) 19.25 sq.cm
c) 44 sq.cm
d) 77 sq.cm
Question 4: What is the ratio of radius of incircle and circumcircle for an equilateral triangle with an area of $22\surd3$ square units?
a) 1/2
b) 1/4
c) 1/3
d) 2/3
Question 5: If the radius (r) of a circle is increased by ‘x’ units, what is the number of units by which the circumference of the circle is increased?
a) $\pi$
b) $2\pi$
c) $2\pi r$
d) $2\pi x$
Instructions
Consider the following information and questions based on it.
P, W, Q, X, R, Y, Z and S sit randomly in a circle facing each other.
1. Y sits exactly between Q and R.
2. P does not sit next to either X or Z.
3. Z is to the immediate right of Q.
4. S sits third from the left of X and the right of Y.
Question 6: If they all were facing outside the circle, W would be to the left of
a) X
b) S
c) P
d) Cannot be determined
Question 7: If the area of a circle is $9\pi$ sq. cm then its circumference is
a) 9 cm
b) $6 \pi$ cm
c) $3 \pi$ cm
d) 6 cm
Question 8: In the circle below, chord $\overline{AB}$ is extended to meet the tangent $\overline{DE}$ at D. If $\overline{AB}$ = 9 cm and $\overline{BD}$ = 3 cm, find the length of $\overline{DE}$.
a) 5 cm
b) 4 cm
c) $\sqrt{27}$ cm
d) 6 cm
Question 9: The area of a circle is 616 sq.m. Find its diameter. $(\pi=22/7)$
a) 7m
b) 14m
c) 28m
d) 56m
Question 10: A circle touches all the sides of a quadrilateral PQRS. whose sides, PQ = 2cm, QR = 3cm and RS = 4cm, What is the length of PS?
a) 3 cm
b) 2 cm
c) 1 cm
d) 4 cm
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Question 11: The area of the circumcircle of a right angle triangle whose sides are given as 2 cm, $2\sqrt{3}$ cm and 4 cm, is given by
a) $6 \pi$ $cm^2$
b) $4 \pi$ $cm^2$
c) $16 \pi$ $cm^2$
d) $12 \pi$ $cm^2$
Question 12: A copper wire when bent in the form of a square encloses an area of $121\ cm^2$. If the same wire is bent into the form of a circle, find the area of the circle:(Use:$\pi=\frac{22}{7}$)
a) $154\ cm^2$
b) $153\ cm^2$
c) $155\ cm^2$
d) $150\ cm^2$
Question 13: The area of a circle got increased by 22 cm when the radius was increased by 1 cm. What was the original radius?
a) 5 cm
b) 9 cm
c) 3 cm
d) 7 cm
Question 14: If a chord of length 24 cm is at a distance of 5 cm from centre, then find the radius of the circle.
a) 17 cm
b) 15 cm
c) 25 cm
d) 13 cm
Question 15: What is the area of a circular track that runs around a circle with circumference 440 m, having a width of 7 m?
a) 3856 $m^2$
b)
3234 .
c) 3900 $m^2$
d) 3204 $m^2$
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Answers & Solutions:
1) Answer (A)
The area of the arc = $ \frac{36}{360} * \pi * r^2$ = 3.85
Solving, we get r = 3.5
Length of the arc = $ \frac{36}{360} * 2 * \pi * r$ = 2.2
2) Answer (B)
Let say, radius of the circle is r.
Area of the square =484 sq. cm.
So,Side of the square$\sqrt{484\ }=22$ cm.
same wire is used to form the circle.
So, perimeter of the circle will be same as perimeter of the square.
So, $2\times\pi\ \times r=4\times\ 22.$
or,$r=\frac{88\ }{2\pi\ }=14.$
So,Area of the circle$\pi r^2=\pi\times14^2=\ \frac{\ 22}{7}\times14^2=616\ cm^2.\ $
B is correct choice.
3) Answer (B)
According to question :
2×(22/7)×r=22.
or,r=(7/2)=3.5.
So,Area of circle=$πr^2$
=$(22/7)(3.5)^2$.
=38.5 $cm^2$
So,Area of semicircle = (38.5/2)=19.25$cm^2$
B is correct choice.
4) Answer (A)
We know that
radius of incircle of equilateral triangle = $\frac{side}{2\sqrt{3}}$
radius of circumcircle of equilateral triangle = $\frac{side}{\sqrt{3}}$
hence there ratio = 1:2
5) Answer (D)
Circumference of circle = $2\pi\ r$
Circumference of circle when radius increased by x = $2\pi\ \left(r+x\right)=2\pi\ r+2\pi\ x$
Hence, circumference of the circle increased by $2\pi x$ units.
6) Answer (C)
W seats left of P.
7) Answer (B)
$\pi\ r^2=9\pi\ $
$r^2=9$
r = 3
Circumference = $2\pi\ r$= $2\pi\ \times\ 3=6\pi\ $
8) Answer (D)
9) Answer (C)
Area of the circle = $\pi\ r^2$
616 = $\frac{22}{7}\times\ r^2$
$r^2=196$
r = 14 m
Hence diameter = 2r = 28 m
10) Answer (A)
11) Answer (B)
C which is the length of the hypotenuse of a right angled triangle of a circumcircle passing through the centre of the circumcircle.
Sides are given as 2 cm, $2\sqrt{3}$ cm and 4 cm,
Since hypotenuse of a right angled triangle is the longest side i.e c = 4 cm which makes it the diameter of the circumcircle.
Therefore, area of the circumcircle = $\pi \times R^2 = \pi \times 2^2$
$=\pi \times 4 cm^2$
Option B is correct.
12) Answer (A)
Given
Encloses area by forming the square = 121$cm^2$
we know that
Area of square =$a^2$
so,
$a^2$=121
a=11 cm
now perimeter of square = perimeter of circle according to question
from here we get r=7 cm
Area of circle =$\pi\text{r}^2$
= $\frac{22}{7}\cdot7^2=154cm^2$ Ans
13) Answer (C)
area of circle = $\pi\times r^2$
radius = r
ATQ
$\pi(r + 1)^2- \pi\times r^2 = 22$
$\pi((r + 1)^2- r^2) = 22$
$\frac{1+ 2r}{7}=1$
2r = 6
r = 3cm
14) Answer (D)
it becomes a right angle triangle
chord = 24 cm
so
base of triangle = half of 24 = 12 cm
height = 5cm
apply pythagoras’s thearom
hypotenuse = $\sqrt(5^2 + 12^2)$
= 13cm
15) Answer (B)
Circumference=$ 2\times \frac{22}{7} \times R=440$
so Radius of circle=70
now radius of track=7 so RADIUS of bigger circle=77.
AREA of TRACK = AREA OF BIGGER CIRCLE – AREA OF SMALL CIRCLE
=$ \frac {22}{7} \times 77 \times 77 – \frac {22}{7} \times 70 \times 70$
=3234m
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We hope this Top-15 Circle Questions pdf for RRB NTPC exam will be highly useful for your Preparation.
